1 + x + x 2 + x 3 + x 4 + … = 1 5 , x = ?
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First note that this is an infinite geometric series so | x | < 1, so x = 1 4 1 5 is clearly not the answer. Now: 1 + x + x 2 + x 3 + x 4 + . . . . = 1 − x 1 = 1 5 ⇒ x = 1 − 1 5 1 so x = 1 5 1 4
It is not given in question that |x|<1
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Well, it's not given explicitly, but it's obvious after 10 seconds or less of consideration ...
The right-hand side is given as finite so the geometric series on the left hand side must be convergent so |x| < 1.
as x+x^2+x^3+x^4...=14.....(i) take x common so it will become 1+x+x^2+x^3+x^4...=14/x...(ii) also we can write the equation (ii) as 1+(x+x^2+x^3+x^4...)=14/x...(iii) substituting equation (i) in equation (iii) will become 1+14=14/x ==> x=14/15
we know that x + x^2 + x^3........=14 Now, take x common. Then x(1 + x + x^2 + x^3 + x^4.....) = 14 x(1+14)=14 x(15) = 14 so, x = 14/15
It is a geometric series with sum x/(1-x) for x between 0 and 1.
Then the equation can be written as: x/(1-x) = 14
From which follows that x = 14/15
Qed
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1 + x + x 2 + x 3 + x 4 + . . . = 1 5 x ( 1 + x + x 2 + x 3 + x 4 + . . . ) = 1 4 x ( 1 + ( x + x 2 + x 3 + x 4 + . . . ) ) = 1 4 x ( 1 + 1 4 ) = 1 4
x = 1 5 1 4